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The coefficient of linear expansion of crystal in one direction is ${\alpha _1}$ and that in every direction perpendicular to it is ${\alpha _2}$. The coefficient of cubical expansion is
${\alpha _1} + {\alpha _2}$
$2{\alpha _1} + {\alpha _2}$
${\alpha _1} + 2{\alpha _2}$
None of these
Solution
(c) $V = {V_0}(1 + \gamma \Delta \theta )$
${L^3} = {L_0}(1 + {\alpha _1}\Delta \theta )L_0^2{(1 + {\alpha _2}\Delta \theta )^2}$$ = L_0^3(1 + {\alpha _1}\Delta \theta )\,{(1 + {\alpha _2}\Delta \theta )^2}$
Since $L_0^3 = {V_0}$ and ${L^3} = V$
Hence $1 + \gamma \Delta \theta = (1 + {\alpha _1}\Delta \theta )\,{(1 + {\alpha _2}\Delta \theta )^2}$
$\tilde = \,(1 + {\alpha _1}\Delta \theta )\,(1 + 2{\alpha _2}\Delta \theta )$ $\tilde = \,(1 + {\alpha _1}\Delta \theta + 2{\alpha _2}\Delta \theta )$
==> $\gamma = \alpha_1 + 2\alpha_2 $
